Sample Exercise 3.1 Interpreting and Balancing Chemical Equations

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1 Sample Exercise 3.1 Interpreting and Balancing Chemical EquationsThe following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the bluespheres are nitrogen atoms. (a) Write the chemical formulas for the reactants and products. (b) Write a balancedequation for the reaction. (c) Is the diagram consistent with the law of conservation of mass?Solution(a) The left box, which represents reactants, contains two kinds of molecules, those composed of two oxygen atoms (O2) and those composed of one nitrogen atom and one oxygen atom (NO). The right box, which represents products, contains only one kind of molecule, which is composed of one nitrogen atom and two oxygen atoms (NO2).(b) The unbalanced chemical equation is O2 + NO → NO2 (unbalanced)An inventory of atoms on each side of the equation shows that there are one N and three O on the left side of the arrow and one N and two O on the right. To balance O we must increase the number of O atoms on the right while keeping the coefficients for NO and NO2 equal. Sometimes a trial-and-error approach is required; we need to go back and forth several times from one side of an equation to the other, changing coefficients first on one side of the equation and then the other until it is balanced. In our present case, let’s start by increasing the number of O atoms on the right side of the equation by placing the coefficient 2 in front of NO2:O2 + NO → 2 NO2 (unbalanced)

2 Sample Exercise 3.1 Interpreting and Balancing Chemical EquationsContinuedNow the equation gives two N atoms and four O atoms on the right, so we go back to the left side. Placing thecoefficient 2 in front of NO balances both N and O:(c) The reactants box contains four O2 and eight NO. Thus, the molecular ratio is one O2 for each two NO, asrequired by the balanced equation. The products box contains eight NO2, which means the number of NO2product molecules equals the number of NO reactant molecules, as the balanced equation requires.There are eight N atoms in the eight NO molecules in the reactants box. There are also 4 × 2 = 8 O atoms inthe O2 molecules and 8 O atoms in the NO molecules, giving a total of 16 O atoms. In the products box, we findeight NO2 molecules, which contain eight N atoms and 8 × 2 = 16 O atoms. Because there are equal numbers ofN and O atoms in the two boxes, the drawing is consistent with the law of conservation of mass.Practice Exercise 1In the following diagram, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms.

3 Sample Exercise 3.1 Interpreting and Balancing Chemical EquationsContinuedThe two reactants combine to form a single product, ammonia, NH3, which is not shown. Write a balancedchemical equation for the reaction. Based on the equation and the contents of the left (reactants) box, find howmany NH3 molecules should be shown in the right (products) box. (a) 2, (b) 3, (c) 4, (d) 6, (e) 9.Practice Exercise 2In the following diagram, the white spheres represent hydrogen atoms, the black spheres carbon atoms, and the redspheres oxygen atoms.In this reaction, there are two reactants, ethylene, C2H2, which is shown, and oxygen, O2, which is not shown, and two products, CO2 and H2O, both of which are shown. (a) Write a balanced chemical equation for the reaction. (b) Determine the number of O2 molecules that should be shown in the left (reactants) box.

4 Sample Exercise 3.2 Balancing Chemical EquationsBalance the equationNa(s) + H2O (l) → NaOH(aq) + H2(g)SolutionBegin by counting each kind of atom on the two sides of the arrow. There are one Na, one O, and two H on the leftside, and one Na, one O, and three H on the right. The Na and O atoms are balanced, but the number of H atoms is not.To increase the number of H atoms on the left, let’s try placing the coefficient 2 in front of H2O:Na(s) + 2H2O (l) → NaOH(aq) + H2(g)Although beginning this way does not balance H, it does increase the number of reactant H atoms, which we need todo. (Also, adding the coefficient 2 on H2O unbalances O, but we will take care of that after we balance H.) Now thatwe have 2 H2O on the left, we balance H by putting the coefficient 2 in front of NaOH:Na(s) + 2H2O (l) → 2 NaOH(aq) + H2(g)Balancing H in this way brings O into balance, but now Na is unbalanced, with one Na on the left and two on the right.To rebalance Na, we put the coefficient 2 in front of the reactant:2 Na(s) + 2H2O (l) → 2 NaOH(aq) + H2(g)We now have two Na atoms, four H atoms, and two O atoms on each side. The equation is balanced.

5 Sample Exercise 3.2 Balancing Chemical EquationsContinuedComment Notice that we moved back and forth, placing a coefficient in front of H2O, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side and then in front of a formula on the other side until the equation is balanced. You can always tell if you have balanced your equation correctly by checking that the number of atoms of each element is the same on the two sides of the arrow, and that you’ve chosen the smallest set of coefficients that balances the equation.Practice Exercise 1The unbalanced equation for the reaction between methane and bromine is__CH4(g) + __ Br2(l) → __ CBr4 (s) + __ HBr(g)Once this equation is balanced what is the value of the coefficient in front of bromine Br2?(a) 1, (b) 2, (c) 3, (d) 4, (e) 6.Practice Exercise 2Balance these equations by providing the missing coefficients:(a) __ Fe(s) + __ O2(g) → __Fe2O3(s)(b) __ Al(s) + __ HCl(aq) → __ AlCl3(aq) + __ H2(g)(c) __ CaCO3(s) + __ HCl(aq) → __ CaCl2(aq) + __ CO2(g) + __ H2O(l)

6 Sample Exercise 3.3 Writing Balanced Equations for Combination and Decomposition ReactionsWrite a balanced equation for (a) the combination reaction between lithium metal and fluorine gas and (b) thedecomposition reaction that occurs when solid barium carbonate is heated (two products form, a solid and a gas).Solution(a) With the exception of mercury, all metals are solids at room temperature. Fluorine occurs as a diatomic molecule. Thus, the reactants are Li(s) and F2(g). The product will be composed of a metal and a nonmetal, so we expect it to be an ionic solid. Lithium ions have a 1+ charge, Li+, whereas fluoride ions have a 1− charge, F−. Thus, the chemical formula for the product is LiF. The balanced chemical equation is2 Li(s) + F2(g) → 2 LiF(s)(b) The chemical formula for barium carbonate is BaCO3. As mentioned, many metal carbonates decompose to metaloxides and carbon dioxide when heated. In Equation 3.7, for example, CaCO3 decomposes to form CaO and CO2.Thus, we expect BaCO3 to decompose to BaO and CO2. Barium and calcium are both in group 2A in the periodictable, which further suggests they react in the same way:BaCO3(s) → BaO(s) + CO2 (g)Practice Exercise 1Which of the following reactions is the balanced equation that represents the decomposition reaction that occurs when silver (I) oxide is heated? (a) AgO(s) → Ag(s) + O(g); (b) 2 AgO(s) → 2 Ag(s) + O2(g); (c) Ag2O(s) → 2 Ag(s) + O2(g); (d) Ag2O(s) → 4 Ag(s) + O2 (g); (e) Ag2O(s) → 4 Ag(s) + O 2(g);

8 Sample Exercise 3.4 Writing Balanced Equations for Combustion ReactionsWrite the balanced equation for the reaction that occurs when methanol, CH3OH(l), is burned in air.SolutionWhen any compound containing C, H, and O is combusted, it reacts with the O2(g) in air to produce CO2(g) and H2O(g). Thus, the unbalanced equation is2 CH3OH(l) + O2(g) → CO2(g) + H2O(g)The C atoms are balanced, one on each side of the arrow. Because CH3OH has four H atoms, we place the coefficient 2 in front of H2O to balance the H atoms:CH3OH(l) + O2(g) → CO2(g) + 2 H2O(g)Adding this coefficient balances H but gives four O atoms in the products. Because there are only three O atoms in the reactants, we are not finished. We can place the coefficient in front of O2 to give four O atoms in the reactants

10 Sample Exercise 3.5 Calculating Formula WeightsCalculate the formula weight of (a) sucrose, C12H22O11 (table sugar); and (b) calcium nitrate, Ca(NO3)2Solution(a) By adding the atomic weights of the atoms in sucrose, we find the formula weight to be amu:(b) If a chemical formula has parentheses, the subscript outside the parentheses is a multiplier for all atoms inside. Thus, for Ca(NO3)2 we havePractice Exercise 1Which of the following is the correct formula weight for calcium phosphate? (a) amu, (b) amu, (c) amu, (d) amu, (e) amu.

12 Sample Exercise 3.6 Calculating Percentage CompositionCalculate the percentage of carbon, hydrogen, and oxygen (by mass) in C12H22O11.SolutionLet’s examine this question using the problem-solving steps in the accompanying “Strategies in Chemistry:Problem Solving” essay.Analyze We are given a chemical formula and asked to calculate the percentage by mass of each element.Plan We use Equation 3.10, obtaining our atomic weights from a periodic table. We know the denominator inEquation 3.10, the formula weight of C12H22O11, from Sample Exercise 3.5. We must use that value in threecalculations, one for each element.SolveCheck Our calculated percentages must add up to 100%, which they do. We could have used more significantfigures for our atomic weights, giving more significant figures for our percentage composition, but we haveadhered to our suggested guideline of rounding atomic weights to one digit beyond the decimal point.

14 Sample Exercise 3.7 Estimating Numbers of AtomsWithout using a calculator, arrange these samples in order of increasing numbers of carbon atoms: 12 g 12C, 1 mol C2H2, 9 × 1023 molecules of CO2.SolutionAnalyze We are given amounts of three substances expressed in grams, moles, and number of molecules andasked to arrange the samples in order of increasing numbers of C atoms.Plan To determine the number of C atoms in each sample, we must convert 12 g 12C, 1 mol C2H2, and 9 × 1023molecules CO2 to numbers of C atoms. To make these conversions, we use the definition of mole and Avogadro’snumber.Solve One mole is defined as the amount of matter that contains as many units of the matter as there are C atomsin exactly 12 g of 12C. Thus, 12 g of 12C contains 1 mol of C atoms = 6.02 × 1023 C atoms. One mol of C2H2contains 6.02 × 1023 C2H2 molecules. Because there are two C atoms in each molecule, this sample contains12.04 × 1023 C atoms. Because each CO2 molecule contains one C atom, the CO2 sample contains 9 × 1023 Catoms. Hence, the order is 12 g 12C (6 × 1023 C atoms) < 9 × 1023 CO2 molecules (9 × 1023 C atoms) < 1 molC2H2 (12 × 1023 C atoms).Check We can check our results by comparing numbers of moles of C atoms in the samples because the numberof moles is proportional to the number of atoms. Thus, 12 g of 12C is 1 mol C, 1 mol of C2H2 contains 2 mol C,and 9 × 1023 molecules of CO2 contain 1.5 mol C, giving the same order as stated previously.

16 Sample Exercise 3.8 Converting Moles to Number of AtomsCalculate the number of H atoms in mol of C6H12O6.SolutionAnalyze We are given the amount of a substance (0.350 mol) and its chemical formula C6H12O6. The unknownis the number of H atoms in the sample.Plan Avogadro’s number provides the conversion factor between number of moles of C6H12O6 and number ofmolecules of C6H12O6: 1 mol C6H12O6 = 6.02 × 1023 molecules of C6H12O6. Once we know the number ofmolecules of C6H12O6, we can use the chemical formula, which tells us that each molecule of C6H12O6 contains12 H atoms. Thus, we convert moles of C6H12O6 to molecules of C6H12O6 and then determine the number ofatoms of H from the number of molecules of C6H12O6:Moles C6H12O6 → molecules C6H12O6 → atoms HSolve

17 Sample Exercise 3.8 Converting Moles to Number of AtomsContinuedCheck We can do a ballpark calculation, figuring that 0.35(6 × 1023) is about 2 × 1023 molecules of C6H12O6. We know that each one of these molecules contains 12 H atoms. 12(2 × 1023) gives 24 × 1023 = 2.4 × 1024 H atoms, which is close to our result. Because we were asked for the number of H atoms, the units of our answer are correct. We check, too, for significant figures. The given data had three significant figures, as does our answer.Practice Exercise 1How many sulfur atoms are in (a) 0.45 mol BaSO4 and (b) 1.10 mol of aluminum sulfide?Practice Exercise 2How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b) 1.50 mol of sodium carbonate?

18 Sample Exercise 3.9 Calculating Molar MassWhat is the molar mass of glucose, C6H12O6?SolutionAnalyze We are given a chemical formula and asked to determine its molar mass.Plan Because the molar mass of any substance is numerically equal to its formula weight, we first determine theformula weight of glucose by adding the atomic weights of its component atoms. The formula weight will haveunits of amu, whereas the molar mass has units of grams per mole (g/mol).Solve Our first step is to determine the formula weight of glucose:Because glucose has a formula weight of amu, 1 mol of this substance (6.02 × 1023 molecules)has a mass of g. In other words, C6H12O6 has a molar mass of g/mol.Check A molar mass below 250 seems reasonable based on the earlier examples we have encountered,and grams per mole is the appropriate unit for the molar mass.

19 Sample Exercise 3.9 Calculating Molar MassContinuedPractice Exercise 1A sample of an ionic compound containing iron and chlorine is analyzed and found to have a molar mass of g/mol. What is the charge of the iron in this compound? (a) 1+, (b) 2+, (c) 3+, (d) 4+.Practice Exercise 2Calculate the molar mass of Ca(NO3)2.

20 Sample Exercise 3.10 Converting Grams to MolesCalculate the number of moles of glucose (C6H12O6) in g of C6H12O6.SolutionAnalyze We are given the number of grams of a substance and its chemical formula and asked to calculate thenumber of moles.Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass ofC6H12O6 is g/mol (Sample Exercise 3.9).Solve Using 1 mol C6H12O6 = g C6H12O6 to write the appropriate conversion factor, we haveCheck Because g is less than the molar mass, an answer less than 1 mol is reasonable. The unit mol isappropriate. The original data had four significant figures, so our answer has four significant figures.Practice Exercise 1How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3?

21 Sample Exercise 3.10 Converting Grams to MolesContinuedPractice Exercise 2How many moles of water are in 1.00 L of water, whose density is 1.00 g/mL?

22 Sample Exercise 3.11 Converting Moles to GramsCalculate the mass, in grams, of mol of calcium nitrate.SolutionAnalyze We are given the number of moles and the name of a substance and asked to calculate the number ofgrams in the substance.Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula andatomic weights.Solve Because the calcium ion is Ca2+ and the nitrate ion is NO3−, the chemical formula for calcium nitrate isCa(NO3)2. Adding the atomic weights of the elements in the compound gives a formula weight of amu.Using 1 mol Ca(NO3)2 = g Ca(NO3)2 to write the appropriate conversion factor, we haveCheck The number of moles is less than 1, so the number of grams must be less than the molar mass, g.Using rounded numbers to estimate, we have 0.5 × 150 = 75 g, which means the magnitude of our answer isreasonable. Both the units (g) and the number of significant figures (3) are correct.

24 Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms from Mass(a) How many glucose molecules are in 5.23 g of C6H12O6?(b) How many oxygen atoms are in this sample?SolutionAnalyze We are given the number of grams and the chemical formula of a substance and asked to calculate (a) thenumber of molecules and (b) the number of O atoms in the substance.Plan (a) The strategy for determining the number of molecules in a given quantity of a substance is summarized inFigure We must convert 5.23 g to moles of C6H12O6 and then convert moles to molecules of C6H12O6. Thefirst conversion uses the molar mass of C6H12O6, g, and the second conversion uses Avogadro’s number.Solve Molecules C6H12O6Check Because the mass we began with is less than a mole, there should be fewer than 6.02 × 1023 molecules inthe sample, which means the magnitude of our answer is reasonable. A ballpark estimate of the answer comesreasonably close to the answer we derived in this exercise: 5/200 = 2.5 × 10−2 mol; (2.5 × 10−2)(6 × 1023) = 15 ×1021 = 1.5 × 1022 molecules. The units (molecules) and the number of significant figures (three) are appropriate.

25 Sample Exercise 3.12 Calculating Numbers of Molecules and Atoms from MassContinuedPlan (b) To determine the number of O atoms, we use the fact that there are six O atoms in each C6H12O6 molecule. Thus, multiplying the number of molecules we calculated in (a) by the factor (6 atoms O/1 molecule C6H12O6) gives the number of O atoms.SolveCheck The answer is six times as large as the answer to part (a), exactly what it should be. The number of significant figures (three) and the units (atoms O) are correct.Practice Exercise 1How many chlorine atoms are in 12.2 g of CCl4? (a) 4.77 × 1022, (b) 7.34 × 1024, (c) 1.91 × 1023, (d) 2.07 × 1023.Practice Exercise 2(a) How many nitric acid molecules are in 4.20 g of HNO3?(b) How many O atoms are in this sample?

26 Sample Exercise 3.13 Calculating an Empirical FormulaAscorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?SolutionAnalyze We are to determine the empirical formula of a compound from the mass percentages of its elements.Plan The strategy for determining the empirical formula involves the three steps given in Figure 3.13.Solve(1) For simplicity we assume we have exactly 100 g of material, although any other mass could also be used.In g of ascorbic acid we have g C, 4.58 g H, and g O.(2) Next we calculate the number of moles of each element. We use atomic masses with four significant figures tomatch the precision of our experimental masses.

27 Sample Exercise 3.13 Calculating an Empirical FormulaContinued(3) We determine the simplest whole-number ratio of moles by dividing each number of moles by the smallestnumber of moles.The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to 1 .This suggests we should multiply the ratios by 3 to obtain whole numbers:C : H : O = (3 × 1 : 3 × 1.33:3 × 1) = (3 : 4 : 3)Thus, the empirical formula is C3H4O3.

28 Sample Exercise 3.13 Calculating an Empirical FormulaContinuedCheck It is reassuring that the subscripts are moderate-size whole numbers. Also, calculating the percentage composition of C3H4O3 gives values very close to the original percentages.Practice Exercise 1A g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains g of carbon, g of oxygen, and g of chlorine. What is the empirical formula of this substance? (a) CO2Cl6, (b) COCl2, (c) C0.022O0.022Cl0.044, (d) C2OCl2Practice Exercise 2A g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains g of carbon, g of hydrogen, and g of oxygen. What is the empirical formula of this substance?

29 Sample Exercise 3.14 Determining a Molecular FormulaMesitylene, a hydrocarbon found in crude oil, has an empirical formula of C3H4 and an experimentally determined molecular weight of 121 amu. What is its molecular formula?SolutionAnalyze We are given an empirical formula and a molecular weight of a compound and asked to determine itsmolecular formula.Plan The subscripts in a compound’s molecular formula are whole-number multiples of the subscripts in itsempirical formula. We find the appropriate multiple by using Equation 3.11.Solve The formula weight of the empirical formula C3H4 is3(12.0 amu) + 4(1.0 amu) = 40.0 amuNext, we use this value in Equation 3.11:Only whole-number ratios make physical sense because molecules contain whole atoms. The 3.03 in this casecould result from a small experimental error in the molecular weight. We therefore multiply each subscript in theempirical formula by 3 to give the molecular formula: C9H12.Check We can have confidence in the result because dividing molecular weight by empirical formulaweight yields nearly a whole number.

30 Sample Exercise 3.14 Determining a Molecular FormulaContinuedPractice Exercise 1Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula? (a) C6H, (b) CH2, (c) C5H24, (d) C6H12, (e) C4H8.Practice Exercise 2Ethylene glycol, used in automobile antifreeze, is 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is g/mol. (a) What is the empirical formula of ethylene glycol? (b) What is its molecular formula?

31 Sample Exercise 3.15 Determining an Empirical Formula by Combustion AnalysisIsopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.SolutionAnalyze We are told that isopropyl alcohol contains C, H, and O atoms and are given the quantities of CO2 andH2O produced when a given quantity of the alcohol is combusted. We must determine the empirical formula forisopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample.Plan We can use the mole concept to calculate grams of C in the CO2 and grams of H in the H2O—the masses ofC and H in the alcohol before combustion. The mass of O in the compound equals the mass of the original sampleminus the sum of the C and H masses. Once we have the C, H, and O masses, we can proceed as in SampleExercise 3.13.Solve Because all of the carbon in the sample is converted to CO2, we can use dimensional analysis and thefollowing steps to calculate the mass C in the sample.Using the values given in this example, the mass of C is

32 Sample Exercise 3.15 Determining an Empirical Formula by Combustion AnalysisContinuedBecause all of the hydrogen in the sample is converted to H2O, we can use dimensional analysis and the following steps to calculate the mass H in the sample. We use three significant figures for the atomic mass of H to match the significant figures in the mass of H2O produced.Using the values given in this example, the mass of H isThe mass of the sample, g, is the sum of the masses of C, H, and O. Thus, the O mass isMass of O = mass of sample – (mass of C + mass of H)= g – (0.153 g g) = g OThe number of moles of C, H, and O in the sample is therefore

33 Sample Exercise 3.15 Determining an Empirical Formula by Combustion AnalysisContinuedTo find the empirical formula, we must compare the relative number of moles of each element in the sample, as illustrated in Sample Exercise 3.13.The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C3H8O.Practice Exercise 1The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO2 and g H2O. A separate experiment shows that it has a molar mass of 88.1 g/mol. Which of the following is the correct molecular formula for dioxane? (a) C2H4O, (b) C4H4O2, (c) CH2, (d) C4H8O2.Practice Exercise 2(a) Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO2 and g H2O. What is the empirical formula of caproic acid? (b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

34 Sample Exercise 3.16 Calculating Amounts of Reactants and ProductsDetermine how many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6:C6H12O6(s) + 6 O2(g)→6 CO2(g) + 6 H2O(l)SolutionAnalyze We are given the mass of a reactant and must determine the mass of a product in the given reaction.Plan We follow the general strategy outlined in Figure 3.16:(1) Convert grams of C6H12O6 to moles using the molar mass of C6H12O6.(2) Convert moles of C6H12O6 to moles of H2O using the stoichiometric relationship 1 mol C6H12O mol H2O.(3) Convert moles of H2O to grams using the molar mass of H2O.Solve(1) First we convert grams of C6H12O6 to moles using the molar mass of C6H12O6.(2) Next we convert moles of C6H12O6 to moles of H2O using the stoichiometric relationship mol C6H12O mol H2O.

35 Sample Exercise 3.16 Calculating Amounts of Reactants and ProductsContinued(3) Finally, we convert moles of H2O to grams using the molar mass of H2O.Check We can check how reasonable our result is by doing a ballpark estimate of the mass of H2O. Because the molar mass of glucose is 180 g/mol, 1 g of glucose equals 1/180 mol. Because 1 mol of glucose yields 6 mol H2O, we would have 6/180 = 1/30 mol H2O. The molar mass of water is 18 g/mol, so we have 1/30 × 18 = 6/10 = 0.6 g of H2O, which agrees with the full calculation. The units, grams H2O, are correct. The initial data had three significant figures, so three significant figures for the answer is correct.

36 Sample Exercise 3.16 Calculating Amounts of Reactants and ProductsContinuedPractice Exercise 1Sodium hydroxide reacts with carbon dioxide to form sodium carbonate and water:2 NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)How many grams of Na2CO3 can be prepared from 2.40 g of NaOH? (a) 3.18 g, (b) 6.36 g, (c) 1.20 g, (d) g.Practice Exercise 2Decomposition of KClO3 is sometimes used to prepare small amounts of O2 in the laboratory: 2 KClO3(s) → 2 KCl(s) + 3 O2(g). How many grams of O2 can be prepared from 4.50 g of KClO3?

37 2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)Sample Exercise 3.17 Calculating Amounts of Reactants and ProductsSolid lithium hydroxide is used in space vehicles to remove the carbon dioxide gas exhaled by astronauts. The hydroxide reacts with the carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide?SolutionAnalyze We are given a verbal description of a reaction and asked to calculate the number of grams of one reactant that reacts with 1.00 g of another.Plan The verbal description of the reaction can be used to write a balanced equation:2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)We are given the mass in grams of LiOH and asked to calculate the mass in grams of CO2. We can accomplish this with the three conversion steps in Figure The conversion of Step 1 requires the molar mass of LiOH ( = g/mol). The conversion of Step 2 is based on a stoichiometric relationship from the balanced chemical equation: 2 mol LiOH mol CO2. For the Step 3 conversion, we use the molar mass of CO (16.00) = g/mol.Solve

38 Sample Exercise 3.17 Calculating Amounts of Reactants and ProductsContinuedCheck Notice that g LiOH/mol ≈ 24 g LiOH/mol, 24 g LiOH/mol × 2 mol LiOH = 48 g LiOH, and (44 g CO2/mol)/(48 g LiOH) is slightly less than 1. Thus, the magnitude of our answer, g CO2, is reasonable based on the amount of starting LiOH. The number of significant figures and units are also appropriate.Practice Exercise 1Propane, C3H8 (Figure 3.8), is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane? (a) 5.00 g, (b) g, (c) 2.18 g, (d) 3.63 g.Practice Exercise 2Methanol, CH3OH, reacts with oxygen from air in a combustion reaction to form water and carbon dioxide. What mass of water is produced in the combustion of 23.6 g of methanol?

39 Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting ReactantThe most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):N2(g) + 3 H2(g) → 2 NH3(g)How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?SolutionAnalyze We are asked to calculate the number of moles of product, NH3, given the quantities of each reactant, N2 and H2, available in a reaction. This is a limiting reactant problem.Plan If we assume one reactant is completely consumed, we can calculate how much of the second reactant is needed. By comparing the calculated quantity of the second reactant with the amount available, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant.Solve The number of moles of H2 needed for complete consumption of 3.0 mol of N2 is

40 Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting ReactantContinuedBecause only 6.0 mol H2 is available, we will run out of H2 before the N2 is gone, which tells us that H2 is the limiting reactant. Therefore, we use the quantity of H2 to calculate the quantity of NH3 produced:Comment It is useful to summarize the reaction data in a table:Notice that we can calculate not only the number of moles of NH3 formed but also the number of moles of each reactant remaining after the reaction. Notice also that although the initial (before) number of moles of H2 is greater than the final (after) number of moles of N2, H2 is nevertheless the limiting reactant because of its larger coefficient in the balanced equation.

41 Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting ReactantContinuedCheck Examine the Change row of the summary table to see that the mole ratio of reactants consumed and product formed, 2:6:4, is a multiple of the coefficients in the balanced equation, 1:3:2. We confirm that H2 is the limiting reactant because it is completely consumed in the reaction, leaving 0 mol at the end. Because 6.0 mol H2 has two significant figures, our answer has two significant figures.Practice Exercise 1When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction 2 CH3OH(l) + 3 O2(g) →2 CO2(g) + 4 H2O(g), what is the excess reactant and how many moles of it remains at the end of the reaction?(a) 9 mol CH3OH(l), (b) 10 mol CO2(g), (c) 10 mol CH3OH(l), (d) 14 mol CH3OH(l), (e) 1 mol O2(g).Practice Exercise 2(a) When 1.50 mol of Al and 3.00 mol of Cl2 combine in the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s), which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?

42 Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting ReactantThe reaction2 H2(g) + O2(g) → 2 H2O(g)is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g) (each measured to two significant figures). How many grams of water can form?SolutionAnalyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem.Plan To identify the limiting reactant, we can calculate the number of moles of each reactant and compare their ratio with the ratio of coefficients in the balanced equation. We then use the quantity of the limiting reactant to calculate the mass of water that forms.Solve From the balanced equation, we have the stoichiometric relations

43 Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting ReactantContinuedUsing the molar mass of each substance, we calculate the number of moles of each reactant:The coefficients in the balanced equation indicate that the reaction requires 2 mol of H2 for every 1 mol of O2. Therefore, for all the O2 to completely react, we would need 2 × 47 = 94 mol of H2. Since there are only 74 mol of H2, all of the O2 cannot react, so it is the excess reactant, and H2 must be the limiting reactant. (Notice that the limiting reactant is not necessarily the one present in the lowest amount.)We use the given quantity of H2 (the limiting reactant) to calculate the quantity of water formed. We could begin this calculation with the given H2 mass, 150 g, but we can save a step by starting with the moles of H2, 74 mol, we just calculated:

44 Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting ReactantContinuedCheck The magnitude of the answer seems reasonable based on the amounts of the reactants. The units are correct, and the number of significant figures (two) corresponds to those in the values given in the problem statement.Comment The quantity of the limiting reactant, H2, can also be used to determine the quantity of O2 used:The mass of O2 remaining at the end of the reaction equals the starting amount minus the amount consumed:1500 g − 1200 g = 300 g.Practice Exercise 1Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light–emitting diodes and solar cells:Ga(l) + As(s) → GaAs(s)

45 Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)Sample Exercise 3.19 Calculating the Amount of Product Formed from a Limiting ReactantContinuedIf 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction? (a) 4.94 g As, (b) 0.56 g As, (c) 8.94 g Ga, or (d) 1.50 g As.Practice Exercise 2When a 2.00-g strip of zinc metal is placed in an aqueous solution containing 2.50 g of silver nitrate, the reaction isZn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)(a) Which reactant is limiting? (b) How many grams of Ag form? (c) How many grams of Zn(NO3)2 form? (d) How many grams of the excess reactant are left at the end of the reaction?

46 2 C6H12 (l) + 5 O2(g) → 2 H2C6H3O4(l) + 2 H2O(g)Sample Exercise 3.20 Calculating Theoretical Yield and Percent YieldAdipic acid, H2C6H8O4, used to produce nylon, is made commercially by a reaction between cyclohexane (C6H12) and O2:2 C6H12 (l) + 5 O2(g) → 2 H2C6H3O4(l) + 2 H2O(g)(a) Assume that you carry out this reaction with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid, what is the percent yield for the reaction?SolutionAnalyze We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C6H12). We are asked to calculate the theoretical yield of a product H2C6H3O4 and the percent yield if only 33.5 g of product is obtained.Plan(a) The theoretical yield, which is the calculated quantity of adipic acid formed, can be calculated using the sequence of conversions shown in Figure 3.16.(b) The percent yield is calculated by using Equation 3.14 to compare the given actual yield (33.5 g) with the theoretical yield.

47 Sample Exercise 3.20 Calculating Theoretical Yield and Percent YieldContinuedSolve(a) The theoretical yield is(b)Check We can check our answer in (a) by doing a ballpark calculation. From the balanced equation we know that each mole of cyclohexane gives 1 mol adipic acid. We have 25/84 ≈ 25/75 = 0.3 mol hexane, so we expect 0.3 mol adipic acid, which equals about 0.3 × 150 = 45 g, about the same magnitude as the 43.5 g obtained in the more detailed calculation given previously. In addition, our answer has the appropriate units and number of significant figures. In (b) the answer is less than 100%, as it must be from the definition of percent yield.

48 Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)Sample Exercise 3.20 Calculating Theoretical Yield and Percent YieldContinuedPractice Exercise 1If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, Cl2, to form 7.7 g of titanium (IV) chloride in a combination reaction, what is the percent yield of the product? (a) 65%, (b) 96%, (c) 48%, or (d) 86%.Practice Exercise 2Imagine you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron:Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)(a) If you start with 150 g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe? (b) If your actual yield is 87.9 g, what is the percent yield?